I don't know if this is the correct sub-forum for this question. If it isn't, I'm sorry.

I have this expression that expands based on the variables given:

$\displaystyle a(2,2) = (\sum\limits_{i_{1}=0}^{2-1} (2-i_{1}) - 1) + (\sum\limits_{i_{1}=0}^{2-2} (2-i_{1}) - 1)$

$\displaystyle a(3,2) = (\sum\limits_{i_{1}=0}^{3-1} (3-i_{1}) - 1) + (\sum\limits_{i_{1}=0}^{3-2} (3-i_{1}) - 1)$

$\displaystyle a(2,3) = (\sum\limits_{i_{1}=0}^{2-1}\sum\limits_{i_{2}=0}^{2-1} (2-i_{1}) \times (2-i_{2}) - 1) +$ \line

$\displaystyle (\sum\limits_{i_{1}=0}^{2-2}\sum\limits_{i_{2}=0}^{2-1} (2-i_{1}) \times (2-i_{2}) - 1) +$

\line

$\displaystyle (\sum\limits_{i_{1}=0}^{2-2}\sum\limits_{i_{2}=0}^{2-2} (2-i_{1}) \times (2-i_{2}) - 1)$

\line

$\displaystyle a(3,3) = (\sum\limits_{i_{1}=0}^{3-1}\sum\limits_{i_{2}=0}^{3-1} (3-i_{1}) \times (3-i_{2}) - 1) +$ \line

$\displaystyle (\sum\limits_{i_{1}=0}^{3-2}\sum\limits_{i_{2}=0}^{3-1} (3-i_{1}) \times (3-i_{2}) - 1) +$

\line

$\displaystyle (\sum\limits_{i_{1}=0}^{3-2}\sum\limits_{i_{2}=0}^{3-2} (3-i_{1}) \times (3-i_{2}) - 1)$

$\displaystyle a(4,4) = (\sum\limits_{i_{1}=0}^{4-1}\sum\limits_{i_{2}=0}^{4-1}\sum\limits_{i_{3}=0}^{4-1} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1) +$ \line

$\displaystyle (\sum\limits_{i_{1}=0}^{4-2}\sum\limits_{i_{2}=0}^{4-1}\sum\limits_{i_{3}=0}^{4-1} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1) +$

\line

$\displaystyle (\sum\limits_{i_{1}=0}^{4-2}\sum\limits_{i_{2}=0}^{4-2}\sum\limits_{i_{3}=0}^{4-1} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1) +$

\line

$\displaystyle (\sum\limits_{i_{1}=0}^{4-2}\sum\limits_{i_{2}=0}^{4-2}\sum\limits_{i_{3}=0}^{4-2} (4-i_{1}) \times (4-i_{2}) \times (4-i_{3}) - 1)$

What I want to do is generalize this expression, i.e.: $\displaystyle a(x,y)= ?$, but I don't know how to go about this. I think I have to use recursion but I don't see how it should be done.

Any help with this would be appreciated.